What is the extraneous solution to these equations? $\dfrac{x^2 - 4}{x + 2} = \dfrac{3x + 6}{x + 2}$
Explanation: Multiply both sides by $x + 2$ $ \dfrac{x^2 - 4}{x + 2} (x + 2) = \dfrac{3x + 6}{x + 2} (x + 2)$ $ x^2 - 4 = 3x + 6$ Subtract $3x + 6$ from both sides: $ x^2 - 4 - (3x + 6) = 3x + 6 - (3x + 6)$ $ x^2 - 4 - 3x - 6 = 0$ $ x^2 - 10 - 3x = 0$ Factor the expression: $ (x - 5)(x + 2) = 0$ Therefore $x = 5$ or $x = -2$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.